Continuous Subarray Sum
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Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.
Example 1:
Input: [23, 2, 4, 6, 7], k=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.
解决方案:首先,当nums.size()<=1时,无意义,返回false。定义一个nums.size()+1的数组sum,sum[i]表示前1个数的和,如果存在一个连续的且长度至少为2的子数组和为k的倍数,则存在i,j使得sum[j]-sum[i]是k的倍数。
对于判断sum[j]-sum[i]是k的倍数,当k=0时,sum[j]-sum[i]=0,当k≠0时,(sum[j]-sum[i])%k=0。
算法复杂度为O(n^2)。
代码如下:
class Solution {public: bool checkSubarraySum(vector<int>& nums, int k) { int i,j,temp; vector<int>::iterator it1=nums.begin(); //vector<int>::iterator it2=nums.end(); int len=nums.size(); int sum[len+1]; if(len<=1) return false; sum[0]=0; for(i=1;i<=len;i++) { sum[i]=sum[i-1]+*(it1+i-1); } for(i=0;i<=(len-1);i++) { for(j=i+2;j<=len;j++) { temp=sum[j]-sum[i]; if((k==0&&temp==0)||(k!=0&&temp%k==0)) { return true; } } } return false; }};
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